Integration using partial fractionsEdexcel A-Level Further Maths: Revision notes

A rational function whose denominator contains a quadratic factor $ax^2+c$ (with $a,c>0$, so no real roots) can be split into simpler fractions. The numerator over that quadratic factor must be linear: $\frac{px^2+qx+r}{(x+k)(ax^2+c)}=\frac{A}{x+k}+\frac{Bx+C}{ax^2+c}.$ A constant-only numerator over $ax^2+c$ is not general enough. The fraction must be proper, with the degree of the numerator less than that of the denominator.

Multiply through by the denominator to get an identity, then use substitution and comparing coefficients. Example: $\frac{5}{(x+1)(x^2+4)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+4}$ gives $5=A(x^2+4)+(Bx+C)(x+1)$. Put $x=-1$: $5=5A$, so $A=1$. Compare $x^2$: $0=A+B$, so $B=-1$. Compare constants: $5=4A+C$, so $C=1$. So $\frac{5}{(x+1)(x^2+4)}=\frac{1}{x+1}+\frac{1-x}{x^2+4}$. Check by putting $x=0$ in both sides: $\frac54=1+\frac14$.

$\int\frac{f'(x)}{f(x)}\,dx=\ln|f(x)|+c,\qquad\int\frac{1}{x^2+a^2}\,dx=\frac1a\arctan\frac{x}{a}+c.$ The arctan result is in the formulae booklet. Examples: $\int\frac{2x}{x^2+7}\,dx=\ln(x^2+7)+c$ (no modulus needed because $x^2+7>0$), and $\int\frac{1}{x^2+4}\,dx=\frac12\arctan\frac x2+c$. For $\int\frac{1}{ax^2+c}\,dx$ with $a\ne1$, factor out $a$: $\int\frac{1}{4x^2+9}\,dx=\frac14\int\frac{1}{x^2+\frac94}\,dx=\frac14\cdot\frac23\arctan\frac{2x}{3}=\frac16\arctan\frac{2x}{3}+c$.

Split $\frac{Bx+C}{ax^2+c}$ into two integrals: $\int\frac{Bx+C}{ax^2+c}\,dx=\frac{B}{2a}\ln(ax^2+c)+C\int\frac{1}{ax^2+c}\,dx.$ The $x$ term is a log (the numerator is a multiple of the derivative $2ax$); the constant term is an arctan. For example $\int\frac{1-x}{x^2+4}\,dx=\frac12\arctan\frac x2-\frac12\ln(x^2+4)+c$.

Find $\int_0^3\frac{x+21}{(x+1)(x^2+9)}\,dx$. Partial fractions: $x+21=A(x^2+9)+(Bx+C)(x+1)$. At $x=-1$: $20=10A$, so $A=2$. Then $B=-2$ and $C=3$. $\int_0^3\left(\frac{2}{x+1}+\frac{3}{x^2+9}-\frac{2x}{x^2+9}\right)dx=\left[2\ln(x+1)+\arctan\frac x3-\ln(x^2+9)\right]_0^3$. $=(2\ln4+\frac\pi4-\ln18)-(-\ln9)=4\ln2-\ln2+\frac\pi4=3\ln2+\frac\pi4$. Combine the logs with $\ln a-\ln b=\ln\frac ab$ and give exact answers: $\ln 2$ and $\pi$, not decimals.